A 0.50-g oil droplet with charge +5.0 * 10 -9 C is in a vertical u E field. In w

Question

A 0.50-g oil droplet with charge +5.0 * 10 -9 C is in a vertical u E field. In what direction should the u E field point and what magnitude should it have so that the droplet moves at constant speed? A 0.50-g oil droplet with charge +5.0 * 10 -9 C is in a vertical u E field. In what direction should the u E field point and what magnitude should it have so that the droplet moves at constant speed? A 0.50-g oil droplet with charge +5.0 * 10 -9 C is in a vertical u E field. In what direction should the u E field point and what magnitude should it have so that the droplet moves at constant speed?

Explanation / Answer

Solution: The mass of the oil droplet, m = 0.50 g = (0.50 g)*(1kg/1000g) = 5.0*10-4 kg

The net charge on the oil droplet is, q = + 5.0*10-9 C

Since the charge on the oil droplet is positive, and since the positive charge experiences force in the direction of electric field and since the force due to electric field should support the downward weight of the oil droplet, the electric field should point in the vertically UPWARD direction.

The oil droplet would stay at rest or move with constant speed (zero acceleration, a = 0 m/s2) only when the net vertical force acting on it is zero. That is, weight of the droplet (acting downward) should get cancelled by upward force due to electric field. From Newton’s second law of motion,

Felectric – W = m*a

q*E – m*g = 0

qE = m*g

E = m*g/q

E = (5.0*10-4 kg)*(9.81m/s2)/( 5.0*10-9 C)

E = 9.81*105 N/C

Strength of electric field in the vertically upward direction should be, E = 9.81*105 N/C

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